Proof of Squeeze Theorem for Sequences

Question

Can someone tell me if my proof is correct?

Statement: if $a_n\leq c_n \leq b_n$ for all $n\in\mathbb{N}$ and $\lim_{n\to\infty}a_n = L = \lim_{n\to\infty}b_n$, then $\lim_{n\to\infty}c_n$ exists and is equal to $L$.

Proof: We are given $a_n\leq c_n\leq b_n$. Since $a_n$ converges to $L$, then there exists a $N_1$ such that whenever $n\geq N_1$, then $|a_n - L| < \epsilon_1$. Since $b_n$ converges to $L$, then there exists a $N_2$ such that whenever $n\geq N_2$, then $|b_n - L| < \epsilon_2$. Now choose $J = \max(N_1, N_2)$. For $n\geq J$ we thus we have $L-\epsilon_1 < a_n \leq c_n \leq b_n < L + \epsilon_2$.

Here is my question: can I say choose $N_1$ and $N_2$ such that $\epsilon_1 < \epsilon_2$? If that is the case then we have $L-\epsilon_2 < L- \epsilon_1 \leq a_n \leq c_n \leq b_n < L+\epsilon_2$. This implies that for arbitrary $\epsilon_2 >0$ there exists a $J$ such that whenever $n\geq J$, $|c_n -L| < \epsilon_2$.

Answer 1

Since $\epsilon_1$ and $\epsilon_2$ were arbitrarily-chosen positive numbers, there is no way to guarantee that $\epsilon_1<\epsilon_2.$ In fact, since $N_1$ depends on $\epsilon_1$ (not the other way around) and $N_2$ depends on $\epsilon_2$ (not the other way around), then we may not be able to tell anything about a given $\epsilon_j$ from the value of $N_j.$

However, you have something of the right idea. You want to show that $\lim_{n\to\infty}c_n=L,$ yes? That is, you need to show that for any $\epsilon>0,$ there is some $N$ such that $n\ge N\implies\left\lvert c_n-L\right\rvert<\epsilon.$

Take any $\epsilon>0.$ Since $a_n\to L$ and $b_n\to L,$ then there exist $N_1,N_2$ such that the following hold:

• $n\ge N_1\implies|a_n-L|<\epsilon$
• $n\ge N_2\implies|b_n-L|<\epsilon$

Taking $J=\max\{N_1,N_2\},$ then for any $n\ge J$ we have $|a_n-L|<\epsilon$ and $|b_n-L|<\epsilon.$ From this, we can proceed in much the same way that you did, and reach the desired result.

Answer 2

You've got it a little backwards. Before obtaining $N_{1}$ or $N_{2}$ we have to choose $\epsilon_{1}$ and $\epsilon_{2}$. So, if we take $\epsilon_1=\epsilon_2=\epsilon$, we'll get our $N_{1}$ and $N_{2}$, and then taking $J=\max(N_{1},N_{2})$ as you did, we'll end up with $L-\epsilon\leq c_{n}\leq L+\epsilon$ for all $n\geq J$. Since $\epsilon$ was arbitrary, this implies that $c_{n}\to L$.